Suppose \(x_1,x_2,\ldots,x_n\) are \(n\) functions of \(t\). Consider the following system of \(n\) linear ODEs: \[\begin{equation} \overrightarrow{x}'=A(t)\overrightarrow{x}+\overrightarrow{g}(t), \;\;\;\;(51) \end{equation}\] where \(\overrightarrow{x}=[x_1,\; x_2,\cdots,\; x_n]^T\).

Theorem. The linear system (51) has a solution \(\overrightarrow{x}=\overrightarrow{\phi}(t)\) on an interval where \(A(t)\) and \(\overrightarrow{g}(t)\) are continuous. Now consider the corresponding homogeneous system: \[\begin{equation} \overrightarrow{x}'=A(t)\overrightarrow{x}. \;\;\;\;(52) \end{equation}\]
Theorem. If \(\overrightarrow{x_1},\ldots,\overrightarrow{x_n}\) are solutions of (52), then so is \[\overrightarrow{x}=c_1\overrightarrow{x_1}+\cdots+c_n\overrightarrow{x_n},\] for all scalars \(c_1,\ldots,c_n\). (Principle of superposition)

Theorem. \(\overrightarrow{x_1},\ldots,\overrightarrow{x_n}\) are linearly independent solutions of (52) iff \[W(\overrightarrow{x_1},\ldots,\overrightarrow{x_n})=\det[\overrightarrow{x_1},\ldots,\overrightarrow{x_n}]\neq 0\] for some (actually all) value of \(t\).

Theorem. If \(\overrightarrow{x_1},\ldots,\overrightarrow{x_n}\) are linearly independent solutions of (52), then any solution \(\overrightarrow{x}\) of (52) is a linear combination of \(\overrightarrow{x_1},\ldots,\overrightarrow{x_n}\), i.e., \[\overrightarrow{x}=c_1\overrightarrow{x_1}+\cdots+c_n\overrightarrow{x_n},\] for some scalars \(c_1,\ldots,c_n\).

Any \(n\) linearly independent solutions of (52) are called fundamental solutions of (52) and \[\overrightarrow{x}=c_1\overrightarrow{x_1}+\cdots+c_n\overrightarrow{x_n}\] is called the general solution of (52).

How to solve (52) when entries of \(A\) are constants?
Recall that \(x'=\lambda x\) has the general solution \(x=ce^{\lambda t}\). So a solution of (52) may have a similar form \(\overrightarrow{x}=e^{\lambda t}\overrightarrow{v}\) for some constant vector \(\overrightarrow{v}\). Let's verify that. \[\begin{align*} \overrightarrow{x}'=A\overrightarrow{x} &\implies \lambda e^{\lambda t}\overrightarrow{v}=Ae^{\lambda t}\overrightarrow{v}\\ &\implies \lambda \overrightarrow{v}=A\overrightarrow{v}\\ &\implies A\overrightarrow{v}=\lambda\overrightarrow{v}\\ &\implies \overrightarrow{v} \text{ is an eigenvector of \(A\) corresponding to eigenvalue }\lambda.\\ \end{align*}\] Theorem. If \(\overrightarrow{v_1},\ldots,\overrightarrow{v_n}\) are \(n\) linearly independent eigenvectors of \(A\) corresponding to eigenvalues \(\lambda_1,\ldots,\lambda_n\) respectively, then \(e^{\lambda_1t}\overrightarrow{v_1},\ldots,e^{\lambda_nt}\overrightarrow{v_n}\) are \(n\) linearly independent solutions of \[\overrightarrow{x}'=A\overrightarrow{x}\] and the general solution is \[\overrightarrow{x}=c_1e^{\lambda_1t}\overrightarrow{v_1}+\cdots+c_ne^{\lambda_nt}\overrightarrow{v_n},\] for arbitrary scalars \(c_1,\ldots,c_n\).

Verify: \[\begin{align*} \overrightarrow{x}'&=c_1e^{\lambda_1t}\lambda_1\overrightarrow{v_1}+\cdots+c_ne^{\lambda_nt}\lambda_n\overrightarrow{v_n}\\ A\overrightarrow{x}&=c_1e^{\lambda_1t}A\overrightarrow{v_1}+\cdots+c_ne^{\lambda_nt}A\overrightarrow{v_n}\\ &=c_1e^{\lambda_1t}\lambda_1\overrightarrow{v_1}+\cdots+c_ne^{\lambda_nt}\lambda_n\overrightarrow{v_n}. \end{align*}\] Thus \(\overrightarrow{x}'=A\overrightarrow{x}\).

Example. Find the general solution of \(\overrightarrow{x}'=\left[\begin{array}{rr} 1&1\\4&-2\end{array} \right] \overrightarrow{x}\).

Solution. The eigenvalues of the coefficient matrix are \(2\) and \(-3\) with corresponding eigenvectors \(\left[\begin{array}{r}1\\1\end{array} \right]\) and \(\left[\begin{array}{r}1\\-4\end{array} \right]\) respectively (show all the steps). So the general solution is \[\overrightarrow{x}(t)= c_1e^{2t}\left[\begin{array}{r}1\\1\end{array} \right] +c_2e^{-3t}\left[\begin{array}{r}1\\-4\end{array} \right].\\\]
Example. \[\overrightarrow{x}'=\left[\begin{array}{rrr} 1&3&-3\\2&2&-3\\2&4&-5\end{array} \right] \overrightarrow{x}\] Find the general solution.

Solution. The eigenvalues of the coefficient matrix are \(-2,-1\), and \(1\) with corresponding eigenvectors \(\left[\begin{array}{r}1\\1\\2\end{array} \right]\), \(\left[\begin{array}{r}0\\1\\1\end{array} \right]\), and \(\left[\begin{array}{r}1\\1\\1\end{array} \right]\) respectively (show all the steps). So the general solution is \[\overrightarrow{x}(t)=c_1e^{-2t}\left[\begin{array}{r}1\\1\\2\end{array} \right] +c_2e^{-t}\left[\begin{array}{r}0\\1\\1\end{array} \right] +c_3e^{t}\left[\begin{array}{r}1\\1\\1\end{array} \right].\]
Example. Solve the IVP \(\overrightarrow{x}'=\left[\begin{array}{rr} -1&1\\1&-1\end{array} \right] \overrightarrow{x},\; \overrightarrow{x}(0)=\left[\begin{array}{c} 3\\1\end{array} \right]\).

Solution. The eigenvalues of the coefficient matrix are \(0\) and \(-2\) with corresponding eigenvectors \(\left[\begin{array}{r}1\\1\end{array} \right]\) and \(\left[\begin{array}{r}-1\\1\end{array} \right]\) respectively (show all the steps). So the general solution is \[\overrightarrow{x}(t)=c_1\left[\begin{array}{r}1\\1\end{array} \right] +c_2e^{-2t}\left[\begin{array}{r}-1\\1\end{array} \right].\] \[\begin{align*} \overrightarrow{x}(0)=\left[\begin{array}{c}3\\1\end{array} \right] &\implies c_1\left[\begin{array}{r}1\\1\end{array} \right] +c_2\left[\begin{array}{r}-1\\1\end{array} \right]=\left[\begin{array}{c}3\\1\end{array} \right]\\ &\implies c_1-c_2=3,\; c_1+c_2=1\\ &\implies c_1=2,\; c_2=-1. \end{align*}\] So the solution is \[\overrightarrow{x}(t)=2\left[\begin{array}{r}1\\1\end{array} \right] -e^{-2t}\left[\begin{array}{r}-1\\1\end{array} \right].\] Geometric view: \[\overrightarrow{x}(t)=2\left[\begin{array}{r}1\\1\end{array} \right] -e^{-2t}\left[\begin{array}{r}-1\\1\end{array} \right] \implies \left[\begin{array}{r}x_1\\x_2\end{array} \right]=\left[\begin{array}{r}2+e^{-2t}\\2-e^{-2t}\end{array} \right] \implies x_1=2+e^{-2t},\; x_2=2-e^{-2t}.\] Eliminating \(t\), we get \(x_1+x_2=4\), the trajectory of the particle whose planar motion is described by the given IVP.

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